Why does $(1+frac{1}{n})^n$ as $n toinfty = e$ mean the same thing as $(e^x)'$ = $e^x$
$begingroup$
I would like a proof as to why as the limit of $n$ approaches infinity, $(1+frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.
I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.
I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.
By the way, here is the work that I have done so far:
$(e^x)'$ = $e^x$ $Rightarrow$ derivative formula = $lim_{h to 0} frac{f(x+h)- f(x)}{h}$ $Rightarrow$ plug it in $Rightarrow$ $lim_{h to 0} frac{e^{(x+h)} - e^x}{h}$ = $e^x Rightarrow lim_{h to 0} frac{e^x(e^h - 1)}{h}$ = $e^x$ $Rightarrow$ h = $frac{1}{n}$ $Rightarrow$ $n to infty$ $Rightarrow$ $lim_{n to infty}$ $frac{e^x(e^{(1/n)} - 1)}{frac{1}{n}}$ = $e^x$ $Rightarrow$ $lim_{n to infty} (e^{frac{1}{n}} - 1)n = 1$
This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!
algebra-precalculus exponential-function
edited Nov 24 '18 at 13:33
user21820
38.8k543153
asked Nov 22 '18 at 22:36
Freedom EagleFreedom Eagle
162
$endgroup$
|
show 8 more comments
$begingroup$
I would like a proof as to why as the limit of $n$ approaches infinity, $(1+frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.
I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.
I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.
By the way, here is the work that I have done so far:
$(e^x)'$ = $e^x$ $Rightarrow$ derivative formula = $lim_{h to 0} frac{f(x+h)- f(x)}{h}$ $Rightarrow$ plug it in $Rightarrow$ $lim_{h to 0} frac{e^{(x+h)} - e^x}{h}$ = $e^x Rightarrow lim_{h to 0} frac{e^x(e^h - 1)}{h}$ = $e^x$ $Rightarrow$ h = $frac{1}{n}$ $Rightarrow$ $n to infty$ $Rightarrow$ $lim_{n to infty}$ $frac{e^x(e^{(1/n)} - 1)}{frac{1}{n}}$ = $e^x$ $Rightarrow$ $lim_{n to infty} (e^{frac{1}{n}} - 1)n = 1$
This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!
algebra-precalculus exponential-function
edited Nov 24 '18 at 13:33
user21820
38.8k543153
asked Nov 22 '18 at 22:36
Freedom EagleFreedom Eagle
162
$endgroup$
2
$begingroup$
The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
$endgroup$
– 0x539
Nov 22 '18 at 22:38
1
$begingroup$
A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
$endgroup$
– Yuriy S
Nov 22 '18 at 22:39
$begingroup$
math.stackexchange.com/questions/2067849/…
$endgroup$
– amWhy
Nov 22 '18 at 22:51
$begingroup$
You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
$endgroup$
– gimusi
Nov 22 '18 at 22:59
$begingroup$
@gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
$endgroup$
– amWhy
Nov 22 '18 at 23:02
|
show 8 more comments
$begingroup$
I would like a proof as to why as the limit of $n$ approaches infinity, $(1+frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.
I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.
I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.
By the way, here is the work that I have done so far:
$(e^x)'$ = $e^x$ $Rightarrow$ derivative formula = $lim_{h to 0} frac{f(x+h)- f(x)}{h}$ $Rightarrow$ plug it in $Rightarrow$ $lim_{h to 0} frac{e^{(x+h)} - e^x}{h}$ = $e^x Rightarrow lim_{h to 0} frac{e^x(e^h - 1)}{h}$ = $e^x$ $Rightarrow$ h = $frac{1}{n}$ $Rightarrow$ $n to infty$ $Rightarrow$ $lim_{n to infty}$ $frac{e^x(e^{(1/n)} - 1)}{frac{1}{n}}$ = $e^x$ $Rightarrow$ $lim_{n to infty} (e^{frac{1}{n}} - 1)n = 1$
This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!
algebra-precalculus exponential-function
edited Nov 24 '18 at 13:33
user21820
38.8k543153
asked Nov 22 '18 at 22:36
Freedom EagleFreedom Eagle
162
$endgroup$
I would like a proof as to why as the limit of $n$ approaches infinity, $(1+frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.
I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.
I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.
By the way, here is the work that I have done so far:
$(e^x)'$ = $e^x$ $Rightarrow$ derivative formula = $lim_{h to 0} frac{f(x+h)- f(x)}{h}$ $Rightarrow$ plug it in $Rightarrow$ $lim_{h to 0} frac{e^{(x+h)} - e^x}{h}$ = $e^x Rightarrow lim_{h to 0} frac{e^x(e^h - 1)}{h}$ = $e^x$ $Rightarrow$ h = $frac{1}{n}$ $Rightarrow$ $n to infty$ $Rightarrow$ $lim_{n to infty}$ $frac{e^x(e^{(1/n)} - 1)}{frac{1}{n}}$ = $e^x$ $Rightarrow$ $lim_{n to infty} (e^{frac{1}{n}} - 1)n = 1$
This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!
algebra-precalculus exponential-function
algebra-precalculus exponential-function
edited Nov 24 '18 at 13:33
user21820
38.8k543153
asked Nov 22 '18 at 22:36
Freedom EagleFreedom Eagle
162
edited Nov 24 '18 at 13:33
user21820
38.8k543153
asked Nov 22 '18 at 22:36
Freedom EagleFreedom Eagle
162
edited Nov 24 '18 at 13:33
user21820
38.8k543153
edited Nov 24 '18 at 13:33
user21820
38.8k543153
edited Nov 24 '18 at 13:33
user21820
38.8k543153
38.8k543153
asked Nov 22 '18 at 22:36
Freedom EagleFreedom Eagle
162
asked Nov 22 '18 at 22:36
Freedom EagleFreedom Eagle
162
asked Nov 22 '18 at 22:36
Freedom EagleFreedom Eagle
162
162
2
$begingroup$
The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
$endgroup$
– 0x539
Nov 22 '18 at 22:38
1
$begingroup$
A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
$endgroup$
– Yuriy S
Nov 22 '18 at 22:39
$begingroup$
math.stackexchange.com/questions/2067849/…
$endgroup$
– amWhy
Nov 22 '18 at 22:51
$begingroup$
You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
$endgroup$
– gimusi
Nov 22 '18 at 22:59
$begingroup$
@gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
$endgroup$
– amWhy
Nov 22 '18 at 23:02
|
show 8 more comments
2
$begingroup$
The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
$endgroup$
– 0x539
Nov 22 '18 at 22:38
1
$begingroup$
A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
$endgroup$
– Yuriy S
Nov 22 '18 at 22:39
$begingroup$
math.stackexchange.com/questions/2067849/…
$endgroup$
– amWhy
Nov 22 '18 at 22:51
$begingroup$
You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
$endgroup$
– gimusi
Nov 22 '18 at 22:59
$begingroup$
@gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
$endgroup$
– amWhy
Nov 22 '18 at 23:02
2
2
$begingroup$
The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
$endgroup$
– 0x539
Nov 22 '18 at 22:38
$begingroup$
The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
$endgroup$
– 0x539
Nov 22 '18 at 22:38
1
1
$begingroup$
A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
$endgroup$
– Yuriy S
Nov 22 '18 at 22:39
$begingroup$
A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
$endgroup$
– Yuriy S
Nov 22 '18 at 22:39
$begingroup$
math.stackexchange.com/questions/2067849/…
$endgroup$
– amWhy
Nov 22 '18 at 22:51
$begingroup$
math.stackexchange.com/questions/2067849/…
$endgroup$
– amWhy
Nov 22 '18 at 22:51
$begingroup$
You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
$endgroup$
– gimusi
Nov 22 '18 at 22:59
$begingroup$
You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
$endgroup$
– gimusi
Nov 22 '18 at 22:59
$begingroup$
@gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
$endgroup$
– amWhy
Nov 22 '18 at 23:02
$begingroup$
@gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
$endgroup$
– amWhy
Nov 22 '18 at 23:02
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
That limit expression gives the number $e$ itself: for $e^x$ it's
$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$
if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,
$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$
& it would make no essential difference to it, because of $n$ tending to $infty$.
answered Nov 24 '18 at 15:43
AmbretteOrriseyAmbretteOrrisey
55410
$endgroup$
$begingroup$
How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:31
$begingroup$
Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:52
$begingroup$
It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:34
$begingroup$
The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:38
add a comment |
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1 Answer
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$begingroup$
That limit expression gives the number $e$ itself: for $e^x$ it's
$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$
if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,
$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$
& it would make no essential difference to it, because of $n$ tending to $infty$.
answered Nov 24 '18 at 15:43
AmbretteOrriseyAmbretteOrrisey
55410
$endgroup$
$begingroup$
How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:31
$begingroup$
Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:52
$begingroup$
It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:34
$begingroup$
The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:38
add a comment |
$begingroup$
That limit expression gives the number $e$ itself: for $e^x$ it's
$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$
if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,
$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$
& it would make no essential difference to it, because of $n$ tending to $infty$.
answered Nov 24 '18 at 15:43
AmbretteOrriseyAmbretteOrrisey
55410
$endgroup$
$begingroup$
How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:31
$begingroup$
Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:52
$begingroup$
It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:34
$begingroup$
The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:38
add a comment |
$begingroup$
That limit expression gives the number $e$ itself: for $e^x$ it's
$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$
if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,
$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$
& it would make no essential difference to it, because of $n$ tending to $infty$.
answered Nov 24 '18 at 15:43
AmbretteOrriseyAmbretteOrrisey
55410
$endgroup$
That limit expression gives the number $e$ itself: for $e^x$ it's
$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$
if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,
$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$
& it would make no essential difference to it, because of $n$ tending to $infty$.
answered Nov 24 '18 at 15:43
AmbretteOrriseyAmbretteOrrisey
55410
answered Nov 24 '18 at 15:43
AmbretteOrriseyAmbretteOrrisey
55410
answered Nov 24 '18 at 15:43
AmbretteOrriseyAmbretteOrrisey
55410
answered Nov 24 '18 at 15:43
AmbretteOrriseyAmbretteOrrisey
55410
55410
$begingroup$
How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:31
$begingroup$
Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:52
$begingroup$
It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:34
$begingroup$
The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:38
add a comment |
$begingroup$
How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:31
$begingroup$
Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:52
$begingroup$
It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:34
$begingroup$
The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:38
$begingroup$
How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:31
$begingroup$
How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:31
$begingroup$
Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:52
$begingroup$
Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
$endgroup$
– Freedom Eagle
Nov 25 '18 at 0:52
$begingroup$
It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:34
$begingroup$
It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:34
$begingroup$
The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:38
$begingroup$
The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
$endgroup$
– AmbretteOrrisey
Nov 25 '18 at 10:38
add a comment |
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2
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The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
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– 0x539
Nov 22 '18 at 22:38
1
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A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
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– Yuriy S
Nov 22 '18 at 22:39
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math.stackexchange.com/questions/2067849/…
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– amWhy
Nov 22 '18 at 22:51
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You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
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– gimusi
Nov 22 '18 at 22:59
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@gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
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– amWhy
Nov 22 '18 at 23:02